New coffee, pounds of coffee added?

Question

A coffee shop wants a new favor of coffee. How many pounds of coffee worth $6 a pound should be added to 20 pounds of coffee worth $3 a pound to get a mixture worth $4 a pound? 

Answer 1

So with the information you have, you know that:

• Coffee A worth $6/lb.
• 20 lbs of Coffee B worth $3/lb.
• Want a mixture of Coffees A+B worth $4/lb.

Now, this means that you need to figure out how many pounds of coffee A that need to be added to the 20 lbs of coffee B to get a mixture worth $4/lb.

 

1. Let us assign x = number of lbs of coffee A.
2. Then let us assign y = number of lbs of the mixture of coffees A+B
3. Now, in an equation, you want to set it up so that you have amount of coffee A (x) multiplied by $6/lb plus 20lbs of coffee B multiplied by $3/lb to equal amount of mixture (y) multiplied by $4/lb
4. So this equation looks like:  $6*x + $3*20 = $4*y
5. Then you know that y = 20 + x so substitute that into your equation to get: $6*x + $3*20 = $4*(x + 20)
6. When you distribute on the right hand side, you get: $6*x + $3*20 = $4*x + $4*20
7. Now, I'm going to take out the $ and multiply the numbers to simplify the equation to get: 6x + 60 = 4x + 80
8. Move the x values to the left side (so subtract 4x from both sides) and move the numerical values to the right side (subtract 60 from each side) to get: 6x-4x = 80-60
9. Then you get 2x=20
10. So x=10
11. x=amount of coffee A you need to make the mixture and if x=10, then you need 10lbs of coffee A!